Question

The cartesian equation of a line are $3x-1=6y+2=1-z$. Find the direction ratios and write its equation in vector form.

Answer 1

Given equations of the line are: $3x-1=6y+2=1-z$

Rewriting the above equation, we have,

$\begin{array}{c}3\left(x-\frac{1}{3}\right)=6\left(y+\frac{2}{6}\right)=-\left(z-1\right)\\ \frac{\left(x-\frac{1}{3}\right)}{\frac{1}{3}}=\frac{\left(y+\frac{1}{3}\right)}{\frac{1}{6}}=\frac{\left(z-1\right)}{-1}......\left(i\right)\end{array}$

Now consider the general equation of the line: $\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}.....\left(2\right)$

Where, l,m and n are the direction ratios of the line and the point (a,b,c) lies on the line.

Compare the equation (1), with the general equation (2)

we have $l=1/3,m=1/6andn=-1$

Also, $a=1/3,b=-1/3andc=1$

This shows that the given line passes through $(1/3,-1/3,1)$

Therefore, the given line passes through the point having

position vector $\overrightarrow{a}=\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}$ and is parallel to the vector $\overrightarrow{b}=\frac{1}{3}\hat{i}+\frac{1}{6}\hat{j}-\hat{k}$

So its vector equation is

$\overrightarrow{r}=\left(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}\right)+\lambda \left(\frac{1}{3}\hat{i}+\frac{1}{6}\hat{j}-\hat{k}\right)$